The area of a rectangle is the product of its dimensions
The dimensions of the rectangle are: and
The area is given as:
Express 125 as 5^3
Apply difference of cubes
The area of a rectangle is:
So, by comparison:
Read more about areas at:
Raul was working on some Solving Systems of Equations problems. When he was checking his solutions, he found two problems with i
ncorrect solutions. Find the errors Raul made on the following two problems. Explain what he did wrong for both of these problems and give the correct answers.
Problem 1:
Solve by substitution
4x + 7y = 1
y = -2x-7
4x + 7(-2x-7) = 1
4x-14x-7 = 1
-10x-7 = 1
-10x = 8
x = 8-10
x = -45
y = -2(-45) -7
y = 85-7
y = 85-355
y = -275
(-45,-275)
Problem 2:
Solve by elimination
5x-2y = -2
4x-3y = 4
-15x + 6y = 6 (multiplied first eq by -3)
+ 8x-6y = 4 (multiplied second eq by 2)
------------------
-7x = 10
x = -107
5(-107)-2y = -2
-507-2y = -2
-2y = -147+507
-2y = 367
y = -187
(-107, -187)
Problem 1:
Solve by substitution
4x + 7y = 1
y = -2x-7
4x + 7(-2x-7) = 1
4x-14x-7 = 1
-10x-7 = 1
-10x = 8
x = 8-10
x = -45
y = -2(-45) -7
y = 85-7
y = 85-355
y = -275
(-45,-275)
Problem 2:
Solve by elimination
5x-2y = -2
4x-3y = 4
-15x + 6y = 6 (multiplied first eq by -3)
+ 8x-6y = 4 (multiplied second eq by 2)
------------------
-7x = 10
x = -107
5(-107)-2y = -2
-507-2y = -2
-2y = -147+507
-2y = 367
y = -187
(-107, -187)
1 answer:
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Answer:
+ 8
Step-by-step explanation:
To rationalise the denominator, multiply the numerator and denominator by the conjugate of the denominator.
The conjugate of - 8 is
+ 8 , then